tc 146 2 BridgeCrossing(ｎ人过桥问题）

SRM 146 2 1000BridgeCrossing

---

Problem Statement

A well-known riddle goes like this: Four people are crossing an old bridge. The bridge cannot hold more than two people at once. It is dark, so they can't walk without a flashlight, and they only have one flashlight! Furthermore, the time needed to cross the bridge varies among the people in the group. For instance, let's say that the people take 1, 2, 5 and 10 minutes to cross the bridge. When people walk together, they always walk at the speed of the slowest person. It is impossible to toss the flashlight across the bridge, so one person always has to go back with the flashlight to the others. What is the minimum amount of time needed to get all the people across the bridge?

In this instance, the answer is 17. Person number 1 and 2 cross the bridge together, spending 2 minutes. Then person 1 goes back with the flashlight, spending an additional one minute. Then person 3 and 4 cross the bridge together, spending 10 minutes. Person 2 goes back with the flashlight (2 min), and person 1 and 2 cross the bridge together (2 min). This yields a total of 2+1+10+2+2 = 17 minutes spent.

You want to create a computer program to help you solve new instances of this problem. Given an int[] times, where the elements represent the time each person spends on a crossing, your program should return the minimum possible amount of time spent crossing the bridge.

Definition

• ClassBridgeCrossing
• MethodminTime
• Parametersvector<int>
• Returnsint
• Method signatureint minTime(vector<int> times)

(be sure your method is public)

Limits

• Time limit (s)2.000
• Memory limit (MB)64

Notes

• In an optimal solution, exactly two people will be sent across the bridge with the flashlight each time (if possible), and exactly one person will be sent back with the flashlight each time. In other words, in an optimal solution, you will never send more than one person back from the far side at a time, and you will never send less than two people across to the far side each time (when possible).

Constraints

• times will have between 1 and 6 elements, inclusive.
• Each element of times will be between 1 and 100, inclusive.

Test cases

1.  

   • times{ 1, 2, 5, 10 }

    

   Returns17

    

   The example from the text.
2.  

   • times{ 1, 2, 3, 4, 5 }

    

   Returns16

    

   One solution is: 1 and 2 cross together (2min), 1 goes back (1min), 4 and 5 cross together (5min), 2 goes back (2min), 1 and 3 cross together (3min), 1 goes back (1min), 1 and 2 cross together (2min). This yields a total of 2 + 1 + 5 + 2 + 3 + 1 + 2 = 16 minutes spent.
3.  

   • times{ 100 }

    

   Returns100

    

   Only one person crosses the bridge once.
4.  

   • times{ 1, 2, 3, 50, 99, 100 }

    

   Returns162

    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <typeinfo>
    #include <fstream>
   
    using namespace std;
    int ti[] , tot =  ;
    int a[][] ;
    class BridgeCrossing {
        public:
        int minTime(vector<int> tim) {
            tot =  ;
            for (int i : tim) ti[tot ++] = i ; tot -- ;
            sort (ti +  , ti + tot + ) ;
           // printf ("tot = %d\n" , tot ) ;
           // for (int i = 1 ; i <= tot ; i ++) printf ("%d " , ti[i]) ; puts ("") ;
            if (tot == ) return ti[] ;
            if (tot == ) return ti[] ;
            int sum = ti[] ;
            for (int i = tot ; i >=  ; i -= ) {
                if (i == ) {
                    sum += ti[] + ti[] ;
                }
                else {
                    if (ti[] + ti[i - ] <  * ti[]) sum += ti[] + ti[i] + ti[] + ti[i - ] ;
                    else sum += ti[] + ti[i] + ti[] + ti[] ;
                }
            }
            return sum ;
        }
    };
   
    // CUT begin
    ifstream data("BridgeCrossing.sample");
   
    string next_line() {
        string s;
        getline(data, s);
        return s;
    }
   
    template <typename T> void from_stream(T &t) {
        stringstream ss(next_line());
        ss >> t;
    }
   
    void from_stream(string &s) {
        s = next_line();
    }
   
    template <typename T> void from_stream(vector<T> &ts) {
        int len;
        from_stream(len);
        ts.clear();
        for (int i = ; i < len; ++i) {
            T t;
            from_stream(t);
            ts.push_back(t);
        }
    }
   
    template <typename T>
    string to_string(T t) {
        stringstream s;
        s << t;
        return s.str();
    }
   
    string to_string(string t) {
        return "\"" + t + "\"";
    }
   
    bool do_test(vector<int> times, int __expected) {
        time_t startClock = clock();
        BridgeCrossing *instance = new BridgeCrossing();
        int __result = instance->minTime(times);
        double elapsed = (double)(clock() - startClock) / CLOCKS_PER_SEC;
        delete instance;
   
        if (__result == __expected) {
            cout << "PASSED!" << " (" << elapsed << " seconds)" << endl;
            return true;
        }
        else {
            cout << "FAILED!" << " (" << elapsed << " seconds)" << endl;
            cout << "           Expected: " << to_string(__expected) << endl;
            cout << "           Received: " << to_string(__result) << endl;
            return false;
        }
    }
   
    int run_test(bool mainProcess, const set<int> &case_set, const string command) {
        int cases = , passed = ;
        while (true) {
            if (next_line().find("--") != )
                break;
            vector<int> times;
            from_stream(times);
            next_line();
            int __answer;
            from_stream(__answer);
   
            cases++;
            if (case_set.size() >  && case_set.find(cases - ) == case_set.end())
                continue;
   
            cout << "  Testcase #" << cases -  << " ... ";
            if ( do_test(times, __answer)) {
                passed++;
            }
        }
        if (mainProcess) {
            cout << endl << "Passed : " << passed << "/" << cases << " cases" << endl;
            int T = time(NULL) - ;
            double PT = T / 60.0, TT = 75.0;
            cout << "Time   : " << T /  << " minutes " << T %  << " secs" << endl;
            cout << "Score  : " <<  * (0.3 + (0.7 * TT * TT) / (10.0 * PT * PT + TT * TT)) << " points" << endl;
        }
        return ;
    }
   
    int main(int argc, char *argv[]) {
        cout.setf(ios::fixed, ios::floatfield);
        cout.precision();
        set<int> cases;
        bool mainProcess = true;
        for (int i = ; i < argc; ++i) {
            if ( string(argv[i]) == "-") {
                mainProcess = false;
            } else {
                cases.insert(atoi(argv[i]));
            }
        }
        if (mainProcess) {
            cout << "BridgeCrossing (1000 Points)" << endl << endl;
        }
        return run_test(mainProcess, cases, argv[]);
    }
    // CUT end

   这个问题大家都很熟悉吧，晚上有５个人过河，只有一盏灯，，他们每个人的过河时间为１，２，３，６，１８（另一组人１，１０，１０，１０，１０）。每次最多两个人过河（过河必须有一个人带着灯），且过河的时间按走得慢的那个人算，求５个人过河所需的最少时间？

   当然这个问题转到程序里自然是变成了ｎ个人过河喽。

   如果你把上面得数据跑了一遍的话，得到２９（另一组４３）的话。那么心里大概就有最佳方案的笼统概念了：

   现在令ｎ个人走过的时间分别为a1,a2,a3,a4,a5，……an;（从小到大）

   那么一开始肯定先是１，２过河。

   接着考虑花时间最多的那两个人（必须考虑他们，这点很重要），我们怎么送他们呢？

   第一种方案：

   　　１回来了，让１把ｎ送过去，然后１回来，再让１把ｎ－１送过去：a1 + an + a1 + a(n-1) ;

   第二种方案：

   　　１回来了，让ｎ，ｎ－１过河，然后２回来，再１，２过河: a1 + an + a2 + a2 ;

   （如果觉得叙述方案时有奇怪的地方，请见怪不怪，总之是为了让每轮送两人，并且每轮结束时保证１，２都在对岸）

   所以这两种方案进行选择时，只要比较：a1 + a(n-1) 和　２× a2 中较小的即可。

   。。。大概就是这样，因为每次送两人，所以总人数为奇数时，还有点细节要考虑。